Q. 164.1( 16 Votes )

The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point B the angle of elevation increases to 60°. Find the height of the tower and the distance of the tower from the point A.

Answer :

Let the height of the tower is = h (m)



Distance of point B from foot of the tower is = (m)


In ∆ADC,


tan 30° =


=


√3 h = 20+ -----------(1)


In ∆DCB,


tan 60° =


√3 =


h = √3 ----------(2)


On substituting value of h from eqn. (2) in eqn. (1)


√3× √3 = 20 +


320+


3-= 20


= 10


Therefore distance of point A from tower is


AC = AB+BC


AC = 20+10 30


Ac = 30 m.


Now substituting value of in eqn. (1)


h = 20+10 30


17.32 m.


Therefore height of tower is 17.32 m.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Trick to learn all Trigonometric Formulae28 mins
Champ Quiz | Trigger on Trigonometry47 mins
Fundas of Trigonometry50 mins
Champ Quiz | Trigonometry Important Questions33 mins
NCERT | Trigonometric Identities52 mins
Champ Quiz | NTSE Trigonometry50 mins
Testing the T- Ratios of Specified Angles57 mins
Foundation | Cracking Previous Year IMO QuestionsFREE Class
NCERT | Imp. Qs. on Trigonometry42 mins
Quiz | Trail of Mixed Questions on Trigonometry59 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses